We know that p 2 2 y p 2 2 2 1 y p 2 1 y d 2 1 and

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We know that p 2 2 ' | y   ± ; p 2 2 ' | 2 1 , y   p 2 1 | y   d 2 1 and can therefore estimate p ³ 2 2 ' | y   ± G " 1 ! g ± 1 G p 2 2 ' | 2 1 g   , y   that is, the average value of p 2 2 ' | 2 1 , y   across Gibbs simulated draws for 2 1 Conclusion: for a two-block Gibbs sampler, we can estimate the marginal likelihood from p ³ y   ± p y | 2 1 ' , 2 2 '   p 2 1 '   p 2 2 '   p 2 1 ' | 2 2 ' , y   G " 1 ! g ± 1 G p 2 2 ' | 2 1 g   , y  
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10 How about 3 blocks? Now we want to estimate the denominator of p y   ± p y | 2 '   p 2 '   / p 2 1 ' , 2 2 ' , 2 3 ' | y   p 2 1 ' , 2 2 ' , 2 3 ' | y   ± p 2 1 ' | y   p 2 2 ' | 2 1 ' , y   p 2 3 ' | 2 1 ' , 2 2 ' , y   p 2 1 ' , 2 2 ' , 2 3 ' | y   ± p 2 1 ' | y   p 2 2 ' | 2 1 ' , y   p 2 3 ' | 2 1 ' , 2 2 ' , y   First term can be estimated as before: p ³ 2 1 ' | y   ± G " 1 ! g ± 1 G p 2 1 ' | 2 2 g   , 2 3 g   , y   Third term p 2 3 ' | 2 1 ' , 2 2 ' , y   is known analytically p 2 1 ' , 2 2 ' , 2 3 ' | y   ± p 2 1 ' | y   p 2 2 ' | 2 1 ' , y   p 2 3 ' | 2 1 ' , 2 2 ' , y   Second term: p 2 2 ' | 2 1 ' , y   ± ; p 2 2 ' | 2 1 ' , 2 3 , y   p 2 3 | 2 1 ' , y   d 2 3 But how do we generate a sample from p 2 3 | 2 1 ' , y   ?
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11 Suppose we do a 2-block Gibbs sampler between 2 2 and 2 3 with 2 1 ' fixed throughout: p 2 2 | 2 1 ' , 2 3 , y   p 2 3 | 2 1 ' , 2 2 , y   The ergodic distribution of 2 3 determined by this Markov chain ( q ± 1,..., Q   is p 2 3 | 2 1 ' , y   p ³ 2 2 ' | 2 1 ' , y   ± Q " 1 ! q ± 1 Q p 2 2 ' | 2 1 ' , 2 3 q   , y   So we estimate p y   from p y | 2 1 ' , 2 2 ' , 2 3 '   p 2 1 '   p 2 2 '   p 2 3 '   p ³ 2 1 ' , 2 2 ' , 2 3 ' | y   p ³ 2 1 ' , 2 2 ' , 2 3 ' | y   ± G " 1 !
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  • Winter '09
  • JamesHamilton
  • Gibbs, Akaike information criterion, Bayesian information criterion, y|, Deviance information criterion

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